Problem: $\dfrac{d}{dx}[\sqrt{\cos(x)}]=\,?$ Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{\sin(x)}{2\sqrt{\cos(x)}}$ (Choice B) B $\dfrac{\cos(x)}{2\sqrt{\sin(x)}}$ (Choice C) C $\dfrac{\sin(x)}{2\sqrt{\cos(x)}}$ (Choice D) D $-\dfrac{\cos(x)}{2\sqrt{\sin(x)}}$
Explanation: Since $\sqrt{\cos(x)}$ is a composite function, we can use the chain rule. The chain rule says: $\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right]={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)}$ To use it, we need to recognize our function as a composite function like $w\big(u(x)\big)$. $\underbrace{\sqrt{~\overbrace{\cos(x)}^{\text{inner}}~}}_{\text{outer}}$ So if $\sqrt{\cos(x)}=w(u(x))$, then: $\begin{aligned} {u(x)}&={\cos(x)} &&\text{inner function} \\\\ w(x)&=\sqrt{x}&&\text{outer function} \end{aligned}$ Before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown, but hopefully these derivatives are familiar by now). $\begin{aligned} {u'(x)}&={-\sin(x)} \\\\ {w'(x)}&={\dfrac{1}{2\sqrt{x}}} \end{aligned}$ Now let's apply the chain rule: $\begin{aligned} \dfrac{d}{dx}[\sqrt{\cos(x)}]&=\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right] \\\\ &={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)} \\\\ &={\dfrac{1}{2\sqrt{({\cos(x)})}}} \cdot {-\sin(x)} \\\\ &=-\dfrac{\sin(x)}{2\sqrt{\cos(x)}} \end{aligned}$